X∼P(λ) and Y∼P(μ) meaning that X and Y are Poisson distributions. What is the probability distribution law of X+Y. I know it is X+Y∼P(λ+μ) but I don’t understand how to derive it.

**Answer**

This only holds if X and Y are independent, so we suppose this from now on. We have for k≥0:

\begin{align*}

P(X+ Y =k) &= \sum_{i = 0}^k P(X+ Y = k, X = i)\\

&= \sum_{i=0}^k P(Y = k-i , X =i)\\

&= \sum_{i=0}^k P(Y = k-i)P(X=i)\\

&= \sum_{i=0}^k e^{-\mu}\frac{\mu^{k-i}}{(k-i)!}e^{-\lambda}\frac{\lambda^i}{i!}\\

&= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \frac{k!}{i!(k-i)!}\mu^{k-i}\lambda^i\\

&= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \binom ki\mu^{k-i}\lambda^i\\

&= \frac{(\mu + \lambda)^k}{k!} \cdot e^{-(\mu + \lambda)}

\end{align*}

Hence, X+ Y \sim \mathcal P(\mu + \lambda).

**Attribution***Source : Link , Question Author : Community , Answer Author : martini*